Question

Efectuați: (1/(2-√3)+2/(√3-1))•(4/(3√2+4)-3/(3-2√2). Sa faceti o poza cu rezolvarea.

Answer 1

   
[tex]\displaystyle\\ \left(\frac{1}{2+ \sqrt{3}}+ \frac{2}{ \sqrt{3}-1}\right) \left(\frac{4}{3\sqrt{2}+4}- \frac{3}{3-2\sqrt{2}}\right) = \text{Rationalizam numitorii}\\\\\\ =\left(\frac{2- \sqrt{3}}{4- 3}+ \frac{2(\sqrt{3}+1)}{ 3-1}\right) \left(\frac{4(3\sqrt{2}-4)}{9\times 2-16}- \frac{3(3+2\sqrt{2})}{9-4\times 2}\right) =\\\\\\ =\left(\frac{2- \sqrt{3}}{1}+ \frac{2\sqrt{3}+2}{ 2}\right) \left(\frac{12\sqrt{2}-16}{18-16}- \frac{9+6\sqrt{2}}{9-8}\right) = [/tex]


[tex]\displaystyle\\ =\left(\frac{^{\b2)}2- \sqrt{3}}{1}+ \frac{2\sqrt{3}+2}{ 2}\right) \left(\frac{12\sqrt{2}-16}{2}- \frac{^{\b2)}9+6\sqrt{2}}{1}\right) =\\\\\\ =\left(\frac{4- 2\sqrt{3}}{2}+ \frac{2\sqrt{3}+2}{ 2}\right) \left(\frac{12\sqrt{2}-16}{2}- \frac{18+12\sqrt{2}}{2}\right) =\\\\\\ =\frac{4- 2\sqrt{3}+2\sqrt{3}+2}{2} \times \frac{12\sqrt{2}-16 - (18+12\sqrt{2})}{2} =\\\\\\ =\frac{4- 2\sqrt{3}+2\sqrt{3}+2}{2} \times \frac{12\sqrt{2}-16 - 18-12\sqrt{2}}{2} = [/tex]


[tex]\displaystyle\\ =\frac{4+2}{2} \times \frac{-16 - 18}{2} = \\\\\\ =\frac{6}{2} \times \frac{-34}{2} = 3\times (-17) = \boxed{\bf -51} [/tex]

Answer 2

Sper sa te ajute. Succes